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Inverse Trigonometric Functions

Short Answer Type

51. Prove that

cos to the power of negative 1 end exponent x minus cos to the power of negative 1 end exponent y equals cos to the power of negative 1 end exponent open square brackets x space y plus square root of left parenthesis 1 minus x squared right parenthesis left parenthesis 1 minus y squared right parenthesis end root close square brackets

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52. Prove that :

tan to the power of negative 1 end exponent open parentheses 1 half close parentheses plus tan to the power of negative 1 end exponent open parentheses 1 over 8 close parentheses equals straight pi over 4

100 Views

53.

Prove space that space tan to the power of negative 1 end exponent 1 fifth plus tan to the power of negative 1 end exponent 1 over 7 plus tan to the power of negative 1 end exponent 1 third plus tan to the power of negative 1 end exponent 1 over 8 equals straight pi over 4.

107 Views

54.

Prove that $tan to the power of negative 1 end exponent space straight t plus tan to the power of negative 1 end exponent fraction numerator 2 space straight l over denominator 1 minus straight t squared end fraction equals tan to the power of negative 1 end exponent fraction numerator 3 straight t minus straight t cubed over denominator 1 minus 3 straight t squared end fraction. straight t greater than 0$

104 Views

55.

Prove that $tan to the power of negative 1 end exponent 1 fourth plus tan to the power of negative 1 end exponent 2 over 9 equals 1 half cos to the power of negative 1 end exponent 3 over 5$

174 Views

56.

Prove that $tan to the power of negative 1 end exponent 1 over 7 plus tan to the power of negative 1 end exponent 2 over 9 equals 1 half cos to the power of negative 1 end exponent 3 over 5$

100 Views

Long Answer Type

57.

Prove that $cot to the power of negative 1 end exponent open parentheses fraction numerator ab plus 1 over denominator straight a minus straight b end fraction close parentheses plus cot to the power of negative 1 end exponent open parentheses fraction numerator bc plus 1 over denominator straight b minus straight c end fraction close parentheses plus cot to the power of negative 1 end exponent open parentheses fraction numerator ca plus 1 over denominator straight c minus straight a end fraction close parentheses equals 0$